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This is probably a very stupid question. I'm sorry.

Let $D$ be a simple normal crossings divisor on some smooth projective variety $D$. By this I mean that the irreducible components $D_i$ are smooth and all possible intersections $\bigcap D_{i_1} \cap \cdots \cap D_{i_k}$ are transversal.

I don't understand what people mean by the singular locus of $D$. In view of the definition, I would say $D$ is smooth. Could anybody help me clarify this point?

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  • $\begingroup$ this is usually called a strict or simple normal crossings divisor $\endgroup$
    – Niels
    Apr 9, 2021 at 20:11

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The points of intersection of distinct reducible components are not smooth. Consider the equation $xy = 0$ in $A^2$, for example. The intersection of the coordinate axes is certainly transversal, but the variety is not smooth at the origin.

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  • $\begingroup$ Ok, so what happens is that the intersections are smooth when considered as subvarieties (in your example the point is smooth) but $D$ is not smooth itself. That's it? $\endgroup$
    – div90
    Jun 20, 2013 at 14:32
  • $\begingroup$ The points on $D$ are smooth as points on the ambient variety; in this example the ambient variety is some affine space, and the divisor $D$ is the union of the x-axis and y-axis. The origin is the only singular point on the divisor $D$ in this example. $\endgroup$
    – Ariyan Javanpeykar
    Jun 20, 2013 at 15:31
  • $\begingroup$ For a divisor with irreducible components having multiplicity one, we can identify it with a reducible subvariety. As a point on this subvariety, a point of intersection is singular, since the dimension of the cotangent space $\frac{m_x}{m_x^2}$ is not equal to the dimension of the subvariety. This is the same reason that an irreducible nodal curve is singular. $\endgroup$
    – Jesse Silliman
    Jun 21, 2013 at 3:18
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In this case, you consider $D$ a divisor (with normal crossing) on a nonsingular variety $X$ of dimension $n$. So, with this notation, the singular set of $D$ is an analitic subvariety of $X$ of codimension $2$. More precisely, if you write $D = \cup_{i} D_{i}$ with $D_{i}$ its irreducible components, the singular set of $D$ is given by

$ sing(D) = \bigcup_{i \neq j} D_{i} \cap D_{j}.$

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